Why does int_a^(ab)1/xdx=int_1^b1/t dt? I can't understand how the integral having limits from a to ab in Step 1 is equivalent to the integral having limits from 1 to b. I'm a beginner here. Please explain in detail. ln(ab)=int_(1)^(ab)1/x dx = int_(1)^(a)(1)/(x)dx+int_(a)^(ab) 1/x dx =int_(1)^(a) 1/x dx +int_(1)^(b)1/(at) d(at)= int_(1)^(a)1/x dx +int_(1)^(b)1/t dt=ln(a)+ln(b)

Why does ${\int }_{a}^{ab}\frac{1}{x}dx={\int }_{1}^{b}\frac{1}{t}dt$?
I can't understand how the integral having limits from $a$ to $ab$ in Step 1 is equivalent to the integral having limits from $1$ to $b$. I'm a beginner here. Please explain in detail.
$\begin{array}{rl}\mathrm{ln}\left(ab\right)={\int }_{1}^{ab}\frac{1}{x}dx& ={\int }_{1}^{a}\frac{1}{x}dx+{\int }_{a}^{ab}\frac{1}{x}dx\\ & ={\int }_{1}^{a}\frac{1}{x}dx+{\int }_{1}^{b}\frac{1}{at}d\left(at\right)\\ & ={\int }_{1}^{a}\frac{1}{x}dx+{\int }_{1}^{b}\frac{1}{t}dt\\ & =\mathrm{ln}\left(a\right)+\mathrm{ln}\left(b\right).\end{array}$
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Claire Larson
The substitution method was already given so take this method:
Let
$f\left(x\right)={\int }_{1}^{x}\frac{dt}{t}$
and
$g\left(x\right)={\int }_{a}^{ax}\frac{dt}{t}$
then we have
${f}^{\prime }\left(x\right)={g}^{\prime }\left(x\right)=\frac{1}{x}$
and
$f\left(1\right)=g\left(1\right)=0$
hence we conclude that
$f\left(x\right)=g\left(x\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }\phantom{\rule{thickmathspace}{0ex}}x\in \mathbb{R}$
Did you like this example?
${\int }_{a}^{ab}\frac{1}{x}dx={\int }_{1}^{b}\frac{1}{t}dt$
Well, it follows by substituting $x=at\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}dx=adt$. Now, the limits in the first integral are $x=a$ to $x=ab$. Hence, if $x=a$, then $t=\frac{a}{a}=1$. and if $x=ab$ , then $t=\frac{ab}{a}=b$ which are your new limits of integration.